CBSE Class 8 Mathematics Part II Chapter 6 Questions and Answers

Class 8 Maths

Algebra Play Class 8 Maths Ganita Prakash Part 2 Chapter 6 NCERT Solutions

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Page 137

Q. Find the dates if the final answers are the following:
(i) 1269   (ii) 394   (iii) 296

Solution:

(i) 1269
Let the month be M and the day be D.
Multiply M by 5: 5M
Add 6: 5M + 6
Multiply by 4: 20M + 24
Add 9: 20M + 33
Multiply by 5: 100M + 165
Add the day: 100M + 165 + D
Given Answer = 1269
1269 = 100M + 165 + D
1269 – 165 = 100M + D
1104 = 100M + D
1100 + 4 = 100M + D
∴ M = 11, D = 4 (i.e. 4th of November)

(ii) 394
Let the month be M and the day be D.
Multiply M by 5: 5M
Add 6: 5M + 6
Multiply by 4: 20M + 24
Add 9: 20M + 33
Multiply by 5: 100M + 165
Add the day: 100M + 165 + D
Given Answer = 394
394 = 100M + 165 + D
394 – 165 = 100M + D
229 = 100M + D
200 + 29 = 100M + D
∴ M = 2, D = 29 (i.e. 29th of February)

(iii) 296
Let the month be M and the day be D.
Multiply M by 5: 5M
Add 6: 5M + 6
Multiply by 4: 20M + 24
Add 9: 20M + 33
Multiply by 5: 100M + 165
Add the day: 100M + 165 + D
Given Answer = 296
296 = 100M + 165 + D
296 – 165 = 100M + D
131 = 100M + D
100 + 31 = 100M + D
∴ M = 1, D = 31 (i.e. 31st of January)

Page 139

Q. Fill the following pyramids:

Solution:

(i)

Let us fill the empty boxes with a, b, c, d, e, and f.
Thus:
a + 22 = 50 → a = 50 – 22 = 28
b + c = a ⇒ b + c = 28 …………. (i)
c + d = 22 …………… (ii)
4 + e = b …………….. (iii)
6 + f = d ……………… (iv)
6 + e = c ……………… (v)

Substituting (iii) and (v) in (i):
(4 + e) + (6 + e) = b + c = 28
10 + 2e = 28
2e = 18 ⇒ e = 9

Substituting (v) and (iv) in (ii):
(6 + e) + (6 + f) = c + d = 22
(6 + 9) + 6 + f = 22
21 + f = 22 ⇒ f = 1

b = 4 + e = 4 + 9 = 13
c = 6 + e = 6 + 9 = 15
d = 6 + f = 6 + 1 = 7

(ii)

Let us fill the empty boxes with a, b, c, d, e, and f.
Thus:
40 + b = a …………. (i)
c + d = 40 …………. (ii)
d + 9 = b …………… (iii)
5 + e = c …………… (iv)
e + 7 = d …………… (v)
7 + f = 9 ⇒ f = 2

Substituting (iv) and (v) in (ii):
(5 + e) + (e + 7) = c + d = 40
2e + 12 = 40
2e = 28 ⇒ e = 14

d = e + 7 = 14 + 7 = 21
b = d + 9 = 21 + 9 = 30
c = 5 + e = 5 + 14 = 19
a = 40 + b = 40 + 30 = 70

(iii)

Let us fill the empty boxes with a, b, c, d, e, and f.
Thus:
a + b = 35 ………… (i)
c + d = a …………… (ii)
d + 7 = b …………… (iii)
3 + 5 = c ⇒ c = 8
5 + e = d …………… (iv)
e + f = 7 ……………. (v)

Substituting (ii) and (iii) in (i):
(c + d) + (d + 7) = a + b = 35
c + 2d + 7 = 35
8 + 2d + 7 = 35 → 2d + 15 = 35
2d = 20 ⇒ d = 10

Substituting d = 10 in (iv):
5 + e = 10 ⇒ e = 5
a = c + d = 8 + 10 = 18
b = d + 7 = 10 + 7 = 17

Substituting e = 5 in (v):
5 + f = 7 ⇒ f = 2

Figure it Out (Page 140)

1. Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

(i)

Solution:
If a = 4, b = 13, and c = 8, then:
Top number = a + 2b + c = 4 + 2(13) + 8 = 4 + 26 + 8 = 38.
Therefore, the number in the topmost row is 38.

(ii)

Solution:
If a = 7, b = 11, and c = 3, then:
Top number = a + 2b + c = 7 + 2(11) + 3 = 7 + 22 + 3 = 32.
Therefore, the number in the topmost row is 32.

(iii)

Solution:
If a = 10, b = 14, and c = 25, then:
Top number = a + 2b + c = 10 + 2(14) + 25 = 10 + 28 + 25 = 63.
Therefore, the number in the topmost row is 63.

2. Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.

Solution:

Let the bottom row be: a, b, c, d.
Second row:
• a + b
• b + c
• c + d

Third row:
• (a + b) + (b + c) = a + 2b + c
• (b + c) + (c + d) = b + 2c + d

Top row:
• (a + 2b + c) + (b + 2c + d) = a + 3b + 3c + d

Therefore, the final expression for the topmost value is a + 3b + 3c + d.

3. Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

Recall the Virahāṅka-Fibonacci number sequence 1, 2, 3, 5, … where each number is the sum of the two numbers before it.

Solution:

(i)

Bottom row: 8, 19, 21, 13 (a = 8, b = 19, c = 21, d = 13)
Top number = a + 3b + 3c + d = 8 + 3(19) + 3(21) + 13 = 8 + 57 + 63 + 13 = 141.

(ii)

Bottom row: 7, 18, 19, 6 (a = 7, b = 18, c = 19, d = 6)
Top number = a + 3b + 3c + d = 7 + 3(18) + 3(19) + 6 = 7 + 54 + 57 + 6 = 124.

(iii)

Bottom row: 9, 7, 5, 11 (a = 9, b = 7, c = 5, d = 11)
Top number = a + 3b + 3c + d = 9 + 3(7) + 3(5) + 11 = 9 + 21 + 15 + 11 = 56.

4. If the first three Virahāṅka-Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahāṅka-Fibonacci numbers?

Solution:
The first three Virahāṅka-Fibonacci numbers are 1, 2, 3.

b = 1 + 2 = 3
c = 2 + 3 = 5
a = b + c = 3 + 5 = 8

Numbers in the grid: 1, 2, 3, 3, 5, 8
Top number: 8
Yes, 1, 2, 3, 3, 5, 8 are all Virahāṅka-Fibonacci numbers.

5. What can you say about the numbers in the pyramid and the number at the top in the following cases?
(i) The first four Virahāṅka-Fibonacci numbers are written in the bottom row of a four-row pyramid.
(ii) The first 29 Virahāṅka-Fibonacci numbers are written in the bottom row of a 29-row pyramid.

Solution:
(i) The first four Virahāṅka-Fibonacci numbers are 1, 2, 3, 5.

d = 1 + 2 = 3
e = 2 + 3 = 5
f = 3 + 5 = 8
b = d + e = 3 + 5 = 8
c = e + f = 5 + 8 = 13
a = b + c = 8 + 13 = 21

The numbers in the pyramid are 1, 2, 3, 5, 3, 5, 8, 8, 13, 21, and the number at the top is 21.

(ii) Top number = (2n − 1)^th term.
For n = 29: 2(29) – 1 = 57^th term.
Therefore, the number at the top is the 57^th number of the Virahāṅka-Fibonacci sequence.

6. If the bottom row of an n-row pyramid contains the first n Virahāṅka-Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top?

Solution:
Each number is obtained by adding the two numbers below it, so all entries are sums of the given Fibonacci numbers.
The top number is the (2n − 1)^th Virahāṅka–Fibonacci number.

Page 142

Q. Create your own calendar trick. For instance, choose a grid of a different size and shape.

Solution:

(i)

Sum of numbers = 1 + 7 + 8 + 9 + 15 = 40.
Let ‘a’ represent the topmost number:

Calendar trick: Sum = a + (a + 6) + (a + 7) + (a + 8) + (a + 14) = 5a + 35.

(ii)

Sum = (10 + 11 + 12) + (17 + 18 + 19) + (24 + 25 + 26) = 162.
Let ‘a’ represent the top-left number:

Calendar trick: Sum = 9a + 72 = 9(a + 8).

(iii)

Sum = 28 + 29 + 30 = 87.
Let ‘a’ represent the left number:

Calendar trick: Sum = a + (a + 1) + (a + 2) = 3a + 3 = 3(a + 1).

(iv)

Sum = 3 + 13 + 23 + 33 + 43 + 21 + 22 + 23 + 24 + 25 = 230.
Let ‘a’ represent the topmost number:

Calendar trick: Sum = 9a + 180 = 9(a + 20).

(v)

Sum = 27 + 37 + 47 + 46 + 48 = 205.
Let ‘a’ represent the topmost number:

Calendar trick: Sum = 5a + 70 = 5(a + 14).

(vi)

Sum = 18 + 19 + 29 + 30 = 96.
Let ‘a’ represent the top-left number:

Calendar trick: Sum = a + (a + 1) + (a + 11) + (a + 12) = 4a + 24 = 4(a + 6).

Q. In the following grids, find the values of the shapes and fill in the empty squares:

Solution:

(i) Let Red circle = C and Blue square = S
2S + C = 27 ⇒ C = 27 − 2S ………… (1)
2C + S = 21
2(27 – 2S) + S = 21
54 – 4S + S = 21 → -3S = -33 ⇒ S = 11
C = 27 − 2(11) ⇒ C = 5
Also: C + S + C = 5 + 11 + 5 = 21

(ii) Let Purple diamond = D and Blue circle = A
A + 2D = 18 ⇒ A = 18 – 2D
D + 2A = 15
D + 2(18 – 2D) = 15
D + 36 – 4D = 15 → -3D = -21 ⇒ D = 7
A = 18 − 14 ⇒ A = 4
Also: D + A + A = 7 + 4 + 4 = 15

Figure it Out (Page 144)

1. Fill the digits 1, 3, and 7 in ☐☐ × ☐ to make the largest product possible.

Solution:
The six possible combinations are: 13 × 7, 17 × 3, 31 × 7, 37 × 1, 71 × 3, 73 × 1.
Comparing the leading candidates:
71 × 3 = (10 × 7 × 3) + (1 × 3) = 210 + 3 = 213
31 × 7 = (10 × 3 × 7) + (1 × 7) = 210 + 7 = 217
Therefore, 31 × 7 gives the largest product (217).

2. Fill the digits 3, 5, and 9 in ☐☐ × ☐ to make the largest product possible.

Solution:
The six possible combinations are: 35 × 9, 39 × 5, 53 × 9, 59 × 3, 93 × 5, 95 × 3.
Comparing the leading candidates:
93 × 5 = (10 × 9 × 5) + (3 × 5) = 450 + 15 = 465
53 × 9 = (10 × 5 × 9) + (3 × 9) = 450 + 27 = 477
Therefore, 53 × 9 gives the largest product (477).

Figure it Out (Page 145 – 146)

1. In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?

Solution:
Suppose the two-digit number is ab and its reverse is ba.
If b > a, difference = (10b + a) – (10a + b) = 9(b – a).
Quotient = 9(b – a) / 9 = (b – a).
Relationship: The quotient is equal to the difference between the two digits of the original number.

2. In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen?
For example: 31 + 13 = 44; 28 + 82 = 110; 12 + 21 = 33. All are divisible by 11. Is this always true? Can we justify this claim using algebra?

Solution:
Yes, this is always true.
Let the number be 10a + b and its reverse be 10b + a.
Sum = (10a + b) + (10b + a) = 11a + 11b = 11(a + b).
Since 11 is a factor, the sum is always divisible by 11.

3. Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3?

Solution:
Sum = (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
= 111a + 111b + 111c
= 111(a + b + c)
Since 111 = 37 × 3:
Sum = 37 × 3(a + b + c).
Therefore, the sum is always divisible by 37, and it is also always divisible by 3.

4. Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works.

Solution:
abcabc = 100000a + 10000b + 1000c + 100a + 10b + c
= 100100a + 10010b + 1001c
= 1001(100a + 10b + c) = 1001 × (abc).
Since 1001 = 7 × 11 × 13:
abcabc / (7 × 11 × 13) = abcabc / 1001 = abc.
You always get back the original 3-digit number.

5. There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?

Solution:
Let starting flowers = x, and flowers placed in each shrine = k.
After pond 1 and shrine 1: 2x – k.
After pond 2 and shrine 2: 2(2x – k) – k = 4x – 3k.
After pond 3 and shrine 3: 2(4x – 3k) = k → 8x – 6k = k → 8x = 7k → x = 7k / 8.
For whole numbers, the smallest positive integer is when k = 8:
x = (7 × 8) / 8 = 7.
Therefore, he started with 7 flowers and placed 8 flowers in each shrine.

6. A farm has some horses and hens. The total number of heads of these animals is 55, and the total number of legs is 150. How many horses and how many hens are on the farm? Can you solve this without letter-numbers?

Solution:
Total animals = 55.
If all 55 animals were hens: Total legs = 55 × 2 = 110.
Extra legs = 150 – 110 = 40.
Each horse has 2 more legs than a hen (4 – 2 = 2).
Number of horses = 40 / 2 = 20 horses.
Number of hens = 55 – 20 = 35 hens.

7. A mother is 5 times her daughter’s age. In 6 years’ time, the mother will be 3 times her daughter’s age. How old is the daughter now?

Solution:
Let daughter's age = x. Mother's age = 5x.
In 6 years: 5x + 6 = 3(x + 6)
5x + 6 = 3x + 18
2x = 12 ⇒ x = 6 years.
The daughter is currently 6 years old.

8. Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, “You have twice as many cows as I do”. Naina says, “That’s true, but if I gave you three of my cows, we would each have the same number of cows”. How many cows do Gauri and Naina have?

Solution:
Let Gauri have x cows and Naina have y cows.
y = 2x
y – 3 = x + 3
2x – 3 = x + 3 ⇒ x = 6.
y = 2(6) = 12.
Therefore, Gauri has 6 cows and Naina has 12 cows.

9. I run a small dosa cart, and my expenses are as follows:
• Rent for the dosa cart is ₹5000 per day.
• The cost of making one dosa (including all the ingredients and fuel) is ₹10.
(i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000?
(ii) If my customers are willing to pay only ₹50 for a dosa, how many dosas should I aim to sell in a day to make a profit of ₹2000?

Solution:
(i) Cost for 100 dosas = 5000 + (100 × 10) = ₹6000.
Required revenue = 6000 + 2000 = ₹8000.
Selling price per dosa = 8000 / 100 = ₹80.

(ii) Let number of dosas = n.
Revenue = 50n.
Cost = 5000 + 10n.
Profit = 50n – (5000 + 10n) = 2000
40n = 7000 ⇒ n = 175 dosas.

10. Evaluate the following sequence of fractions:
1/3 , (1 + 3) / (5 + 7) , (1 + 3 + 5) / (7 + 9 + 11)
What do you observe? Can you explain why this happens?
[Hint: Recall what you know about the sum of the first n odd numbers.]

Solution:
1/3 = 1/3
(1 + 3) / (5 + 7) = 4 / 12 = 1/3
(1 + 3 + 5) / (7 + 9 + 11) = 9 / 27 = 1/3
All fractions simplify to 1/3.
Explanation:
Sum of first n odd numbers = n².
The sum of the next n odd numbers = (Sum of first 2n odd numbers) – (Sum of first n odd numbers) = (2n)² – n² = 4n² – n² = 3n².
Fraction = n² / (3n²) = 1/3.

11. Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie. He heard a voice saying, “I have come to serve you, Oh master”. He woke up and to his surprise, it was a genie!
“Do you want to make money?”, asked the genie. Karim nodded dumbly in bewilderment. The genie continued, “Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double”.
Karim immediately started towards the tree, only to be stopped by the genie. “One moment!”, said the genie. “Since I am bringing you great riches, you should share some of your gains with me. You must give me 8 coins each time you go around the tree.”
Thinking that was a trifling amount, Karim readily agreed.
He went around the tree once. Just as the genie had said, the number of coins in his pocket doubled! He gave 8 coins to the genie. He made another round. Again the number of coins doubled. He gave 8 more coins to the genie. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie!
As Karim began to wonder how the genie tricked him, the genie let out a loud laugh and disappeared.
(i) How many coins did Karim initially have?
(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
(iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim’s coins?